p value Q: In an industrial process it is expected that the

[waxydock]

What is the null hypotheses for this? is it H0: p = 0.1, H1: p > 0.1 ??

In an industrial process it is expected that the proportion of defective
items is 1%. As a quality control measure 100 items are tested on a given day
and 4 out of the sample are found to be defective. Is there evidence that the
population has more than 1% defective items on this day?

(i) Give the distribution of the variables to be observed and state the null and alternative hypotheses of the test.
(ii) State the test statistic for testing the null hypothesis in (i) and explain why it is
an appropriate test statistic.
(iii) Calculate the P-value for the test and draw your conclusion.

 

[galactus]

Be careful with decimals.

The null hypothesis contains the statement of equality.

Here's the way I would interpret it.

\(\displaystyle \L\\\begin{Bmatrix}H_{0}:\overbrace{p\leq{0.01}}^{\text{Claim:\\defects less than\\\underline{or equal to 1%}}}\\H_{a}:\underbrace{p>{0.01}}_{\text{defects not 1%.\\in this case, greater}}\end{Bmatrix}\)


What is your level of significance?. \(\displaystyle {\alpha}=0.01\)?.

The way some problems are stated seems rather ambiguous.

One could also see it as \(\displaystyle H_{0}\) being \(\displaystyle =\) and the \(\displaystyle H_{a}\) being \(\displaystyle \neq\)

 

[waxydock]

Here is my attempt, is this the right way to do it?

H0: P=0.01

H1: P > 0.01

where H0 is the % of defective items expected and H1 is greater than 1% on a given day.

Alternative hypothesis, H1: P < 0.01

for the sample of 100 in now compute the probability,

X ~ B(100, 0.01)
mean 100x0.1 = 10
variance = 100.0.1x0.9 = 9

x ~ N(10, 9)
P(x>=4)
1 -Px<=4)
1-0.99997
=3x10^-5

 

[JakeD]

Here is my attempt, is this the right way to do it?

H0: P=0.01

H1: P > 0.01

where H0 is the % of defective items expected and H1 is greater than 1% on a given day.

Alternative hypothesis, H1: P < 0.01

for the sample of 100 in now compute the probability,

X ~ B(100, 0.01)
mean 100x0.1 = 10 Use 1% = .01, not .1.
variance = 100.0.1x0.9 = 9 Ditto.

x ~ N(10, 9)

Use Z score to get probabilities from normal table.
P(x>=4)
1 -Px<=4)
1-0.99997
=3x10^-5